Fundamentals of Structural Analysis Episode 2 Part 4 ppsx

Fundamentals of Structural Analysis Episode 2 Part 4 ppsx

... = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − +−−−−− −−+−−−−− −− −−−−−−+− −−−−−+ EK L EK C L EK S-EK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L EK L EA CS L EK S L EA C L EK S L EK L EA CS L EK S L EA C EK L EK C L EK SEK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L...

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Fundamentals of Structural Analysis Episode 2 Part 4 doc

... = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − +−−−−− −−+−−−−− −− −−−−−−+− −−−−−+ EK L EK C L EK S-EK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L EK L EA CS L EK S L EA C L EK S L EK L EA CS L EK S L EA C EK L EK C L EK SEK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L...

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Fundamentals Of Structural Analysis Episode 2 Part 4 ppt

... = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − +−−−−− −−+−−−−− −− −−−−−−+− −−−−−+ EK L EK C L EK S-EK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L EK L EA CS L EK S L EA C L EK S L EK L EA CS L EK S L EA C EK L EK C L EK SEK L EK C L EK S L EK C L EK C L EA S L EK L EA CS L EK C L EK C L EA S L EK L EA CS L EK S L...

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Fundamentals of Structural Analysis Episode 2 Part 1 ppsx

... )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-m M F cb = 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m (d) Compute DF at b: 2 m 2 m 4 m a c b 3 kN/m 4 kN 2 m 2 kN 2 m 2 m 4 m a c b 3 kN/m 4 kN 4 ... = − 12 )( )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-m M F cb = 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m (d) Compute DF at b: DF ba : DF b...

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Fundamentals Of Structural Analysis Episode 1 Part 4 ppsx

... shown. (1-a)(1-b) (2) (3) (4- a) (4- b) Problem 2. 4m 4@ 3m=12m 1 kN a b 4m 4@ 3m=12m 1 kN a b 12 kN 4 @ 4m=16 m 2m 3m a b c 6@3m=18m 4m 4m 12 kN a b c 3@3m=9m 4m 4m 15 kN a AB b c 3@3m=9m 4m 4m 15 kN a AB b Truss ... form: x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m...

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Fundamentals of Structural Analysis Episode 1 Part 4 pps

... shown. (1-a)(1-b) (2) (3) (4- a) (4- b) Problem 2. 4m 4@ 3m=12m 1 kN a b 4m 4@ 3m=12m 1 kN a b 12 kN 4 @ 4m=16 m 2m 3m a b c 6@3m=18m 4m 4m 12 kN a b c 3@3m=9m 4m 4m 15 kN a AB b c 3@3m=9m 4m 4m 15 kN a AB b Truss ... form: x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m...

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Fundamentals of Structural Analysis Episode 1 Part 4 pptx

... form: x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 x y 1 2 4m 3m 3 3m 1 2 3 1 kN 1 kN 1 kN 1 kN 1 kN 1 kN Truss Analysis: Force Method, Part ... equation. Contribution of Member Forces Equation Number and Value of Entry Member Number Force Number 2i-1 Coeff. 2i Coeff. 2j-1 Coeff. 2j...

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Fundamentals of Structural Analysis Episode 2 Part 1 pdf

... )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-m M F cb = 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m (d) Compute DF at b: 2 m 2 m 4 m a c b 3 kN/m 4 kN 2 m 2 kN 2 m 2 m 4 m a c b 3 kN/m 4 kN 4 ... = − 12 )( )( 2 Lengthw = − 12 (4) (3) 2 = − 4 kN-m M F cb = 12 )( )( 2 Lengthw = 12 (4) (3) 2 = 4 kN-m (d) Compute DF at b: DF ba : DF b...

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Fundamentals of Structural Analysis Episode 2 Part 2 pptx

... (2) (3) (4) (5) (6) (7) Problem 1. 2 m 2 m 3 m a c b 3 kN/m 4 kN 2EI E I 2 m 2 m 3 m a cb 8 kN 2EI E I 2 kN-m 50 kN ab cd 50 kN 2 m 2 m 2 m 2 m 4 m 50 kN ab c 4 m 4 m 4 m 8 m 2EI 2EI E I E I 2EI2EI 50 ... 0 .22 0 .45 0 1 0 .23 0 .46 0.31 0 MEM M ab M ba M bc M be M eb M ab M ba M bc M be M eb EAM FEM − 62. 5 62. 5 − 62. 5 62. 5 DM 62. 5 62. 5 COM 31.3 31.3 DM...

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Fundamentals of Structural Analysis Episode 2 Part 2 ppt

... kN-m 50 kN ab cd 50 kN 2 m 2 m 2 m 2 m 4 m 50 kN ab c 4 m 4 m 4 m 8 m 2EI 2EI E I E I 2EI2EI 50 kN ab c 4 m 4 m 4 m 8 m E I 2EI2EI 2 m 2 m 2 m 4 kN a b c d 2 m 2 m 2 m 4 kN a b c d d d ... M F − 8 PL 8 PL − ab 2 PL − a 2 bPL − a(1-a)PL a(1-a)PL −(6−8a+3a 2 ) 12 22 wLa (4- 3a) 12 23 wLa − 12 2 wL 12 2 wL − 20 2 wL 12 2 wL − 96 5...

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